Consider the Taylor expansion of the cosine function. We have

${\rm lim}_{s \to \infty} (A_s)=2-{\rm cos}\, x - {\rm cos}\, y$

and

${\rm lim}_{s \to \infty} (B_s)=1- {\rm cos}\, z$ .

For $r>x,y$ the sequence $a_r$ is decreasing as the denominator grows faster than the numerator. Hence for $s>x,y$ the sequence $A_s$ is increasing as $A_{s+4}=A_s+a_{s+2}-a_{s+4}$ and $a_{s+2}>a_{s+4}$ . So if $A_N>0$ for some $N>x,y$ , we have $2-{\rm cos}\, x - {\rm cos}\, y>0$ . Conversely if no such $N$ exists then $A_s \leq 0$ for $s>x,y$ , so its limit, $2-{\rm cos}\, x - {\rm cos}\, y$ , is also less than or equal to $0$ . However as this expression cannot be negative we would have $2-{\rm cos}\, x - {\rm cos}\, y =0$ .

Similarly for $r>z$ the sequence $b_r$ is decreasing, and for $s>z$ the sequence $B_s$ is increasing. So if $B_M>0$ for some $M>z$ we have $1- {\rm cos}\, z>0$ . Conversely if no such $M$ exists then $1- {\rm cos}\, z \leq 0$ . However as this expression cannot be negative we would have $1- {\rm cos}\, z = 0$ .

Note that $2-{\rm cos}\, x - {\rm cos}\, y =0$ precisely when $x,y \in 2\pi \mathbb{Z}$ . Also $1- {\rm cos}\, z = 0$ precisely when $z \in 2\pi \mathbb{Z}$ .

So the analytic form of the theorem may be read:

If for positive reals $x,y,z$ we have $x^n+y^n=z^n$ for some odd integer $n>2$ , then either $x$ or $y$ not in $2\pi \mathbb{Z}$ or $z$ not in $2\pi \mathbb{Z}$ .

Clearly this only fails if for positive integers $a,b,c$ and some odd $n>2$ , we have

$(2\pi a)^n+(2\pi b)^n = (2\pi c)^n$ .

Dividing through by $(2\pi)^n$ we see that $a^n + b^n =c^n$ .

Conversely suppose we have non-zero integers satisfying $a^n + b^n =c^n$ for some $n>2$ . If $n=4k$ we have $(a^k)^4+(b^k)^4=(c^k)^4$ , contradicting example of Fermat's last theorem. Hence if $n$ is even we may replace $a,b,c$ with $a^2,b^2,c^2$ and $n$ with $n/2$ , which will be odd and greater than 1 (and hence greater than 2 as it is odd). So without loss of generality we may assume $n$ odd.

Finally replace $a,b,c$ with their absolute values and if necessary reorder to obtain a positive integer solution. This would be a counterexample to the analytic form of the theorem as stated above.